2-methylpropanol does not form a yellow precipitate with iodine in NaOH because it is a tertiary alcohol, which does not undergo the iodine test reaction typically associated with primary and secondary alcohols. The reaction involves the formation of iodoforms, which requires the presence of a hydrogen atom on the carbon adjacent to the hydroxyl group. Since 2-methylpropanol lacks this structure, it does not produce the necessary iodoform compound, resulting in no yellow precipitate.
The precipitate is copper(II) hydroxide. The chemical reaction is:2 NaOH + CuSO4 = Cu(OH)2 + Na2SO4
The chemical equation is:3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3
The solution of NaOH in methyl orange indicator will turn from yellow to red. Methyl orange is an acid-base indicator that changes color in response to a change in pH. In the presence of a strong base like NaOH, the indicator will change to a red color indicating the basic nature of the solution.
When aqueous solutions of iron(III) sulfate (Fe2(SO4)3) and sodium hydroxide (NaOH) are mixed, a precipitate of iron(III) hydroxide (Fe(OH)3) forms. This occurs due to the reaction between the iron(III) ions and hydroxide ions, leading to the formation of the insoluble hydroxide. The balanced chemical equation for the reaction is: Fe2(SO4)3 + 6 NaOH → 2 Fe(OH)3 (s) + 3 Na2SO4.
Nothing - barium chloride is soluble. You can however precipitate either the barium (e.g. with sodium sulphate, giving barium sulpate, or the chloride, e.g. with silver nitrate giving silver chloride precipitate.
The white precipitate formed when NaOH (aq) is added to MgSO4 (aq) is magnesium hydroxide (Mg(OH)2).
The reaction of ethanol with NaOH and iodine will yield iodoethane (ethyl iodide) as the product. The alcohol group in ethanol will be replaced by the iodine atom in the presence of NaOH.
Copper hydroxide is the precipitate.
Yes, a white precipitate of silver hydroxide (AgOH) will form when solutions of silver nitrate (AgNO3) and sodium hydroxide (NaOH) are mixed. Silver hydroxide is insoluble in water, so it will precipitate out of the solution.
In the starch-iodine test, NaOH is added to create an alkaline environment that allows for the formation of the blue-black complex between starch and iodine. This complex is used as an indicator to detect the presence of starch in a sample.
C6H5COOH + NaOH + I2 -----------> C6H5COOI + NaI + H2O
The chemical equation is:3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3
Iodine solution is commonly used to detect the presence of starch. When iodine solution comes into contact with starch, it changes color from brownish-yellow to a blue-black color. This color change is a positive indication of the presence of starch in a substance.
When mixed with a solution of sodium hydroxide (NaOH), a precipitate will form with magnesium (Mg) cations. The magnesium hydroxide formed is sparingly soluble in water, leading to the precipitation reaction. Potassium (K) and sodium (Na) cations do not precipitate when mixed with NaOH as their hydroxides are highly soluble in water.
The result is a precipitate of Sr(OH)2 as it is insoluble in water. This reaction forms a white precipitate.
Precipitate forms when an INSOLUBLE substance is formed. That means you're looking for ions that form insoluble substances when combined with NaOH and LiCl. That means the ions below are all contenders (use the solubility rules): Ag+ Hg+ Pb2+ Note: PbCl2 is SLIGHTLY soluble in HOT water, but in standard conditions, PbCl2 is considered as insoluble. hope this helps.
The precipitate is copper(II) hydroxide. The chemical reaction is:2 NaOH + CuSO4 = Cu(OH)2 + Na2SO4